pub fn new_birthday_probability(n: u32) -> f64 {
    // TODO: 这里写逻辑
    // todo!()
    let mut probability=1.0;
    let days_in_year=365;

    for i in 0..n {
        probability*=(days_in_year-i) as f64 / days_in_year as f64;
    }

    // 至少两人生日相同的概率
    let res=1.0-probability;

    // 保留四位小数
    (res * 10000.0).round() / 10000.0
}
